# Can you make the triviality of $\langle a,b,c \mid aba^{-1} = b^2, bcb^{-1} = c^2, cac^{-1} = a^2 \rangle$ more trivial?

I recently learned the following pleasant fact. (It was in the proof of Proposition 3.1 of this paper - but don't worry, there's no model theory in this question.)

Let $$G$$ be a group, and let $$a,b,c\in G$$. If $$aba^{-1} = b^2$$, $$bcb^{-1} = c^2$$, and $$cac^{-1} = a^2$$, then $$a = b = c = e$$. Put another way, the group defined by generators and relations $$\langle a,b,c \mid aba^{-1} = b^2, bcb^{-1} = c^2, cac^{-1} = a^2 \rangle$$ is the trivial group.

I came up with the following elementary, but ugly, proof:

The relations can be rewritten as (1) $$ab = b^2a$$, (2) $$bc = c^2b$$, (3) $$ca = a^2c$$.

Using (1), (2), and (3), we can rewrite $$a^4bc = a^4c^2b = c^2ab = c^2b^2a$$.

But we can also rewrite $$a^4bc = b^{16}a^4c = b^{16}ca^2 = c^{2^{16}}b^{16}a^2$$.

So $$c^{2^{16}}b^{16}a^2 = c^2b^2a$$. This implies $$a = b^{-16}c^{2-2^{16}}b^2$$.

Substituting for $$a$$ in (1) above, $$b^{-16}c^{2(1-2^{15})}b^3 = b^{-14}c^{2(1-2^{15})}b^2$$, and cancelling from both sides, $$c^{2(1-2^{15})}b = b^{2}c^{2(1-2^{15})}$$.

But now by (2), we have $$bc^{1-2^{15}} = b^{2}c^{2(1-2^{15})}$$, and $$b = c^{2^{15}-1}$$. But then $$b$$ and $$c$$ commute, so $$bcb^{-1} = c^2$$ implies $$c = c^2$$, and $$c = e$$. It then follows easily that $$a = b = c = e$$.

Question: Is there a better way to see this? i.e. a more abstract proof, or at least one that doesn't involve manipulating words of length $$2^{16}$$?

• A slightly relevant remark, if there are four generators with similar relations, the group will be infinite. That is, $⟨a,b,c,d∣aba^{−1}=b^2, bcb^{−1}=c^2,cdc^{−1}=d^2, dad^{−1}=a^2⟩$ is not finite. c.f. Serre's Trees Page 9. Jan 4, 2019 at 5:02
• ... and now that I can search for an exercise number in a well-known book, I find that there are a number of questions on this site about the same group. For example, Jim Belk gave almost an identical proof to the one I found here, and Martin Brandenburg asked essentially the same question I'm asking here. Together, this evidence makes me think that there's unlikely to be a nicer proof. It would be reasonable to close this question as a duplicate. Jan 4, 2019 at 5:26
• Once you have proved that $a \in \langle b,c \rangle$, you can use the more conceptual argument given by Bhaskar Vashishth in the linked proof, that $G$ is perfect, but $G = \langle b,c \rangle$ is solvable, so $G$ is trivial. Jan 4, 2019 at 15:05
• A quantitative question would go as follows: triviality of $G$ means that one can write, in $F_3$, $a=w$, $w=\prod_{i=1}^ng_ir_ig_i^{-1}$ with $r_i$ among the the three relators. (a) what's the smallest $n$ for this this exists (this is the "area" of the relation)? (c) what is the smallest length $\sum_i(2|g_i|+|r_i|)$ of such a writing? (c) what's the smallest radius (this is, in the loop given by $a^{-1}w$, the largest radius).
– YCor
Jan 11, 2019 at 2:49
• Does this answer your question? Presentation $\langle x,y,z\mid xyx^{-1}y^{-2},yzy^{-1}z^{-2},zxz^{-1}x^{-2}\rangle$ of group equal to trivial group Oct 5, 2022 at 14:47

## 1 Answer

As pointed out by Derek Holt in a comment, if we can show that $$c\in \langle a,b\rangle_G$$, then we can finish the proof in a "conceptual" way:

• Since $$a = [c,a]$$, $$b = [a,b]$$, and $$c = [b,c]$$, $$G' = G$$, i.e., $$G$$ is perfect.
• Since $$c\in \langle a,b\rangle_G$$, $$G = \langle a,b\rangle_G$$, and since $$aba^{-1} = b^2$$, $$G$$ admits a surjective homomorphism from $$H = \langle x,y\mid xyx^{-1} = y^2\rangle$$.
• This group $$H$$ is known as the Baumslag-Solitar group $$\mathrm{BS}(1,2)$$, and is well-known to be solvable. One way to see this is to show that $$H\cong \mathbb{Z}[\frac{1}{2}]\rtimes \mathbb{Z}$$, where $$\mathbb{Z}[\frac{1}{2}]$$ is the additive group of dyadic rationals and $$n\in \mathbb{Z}$$ acts on $$\mathbb{Z}[\frac{1}{2}]$$ by multiplication by $$2^n$$. As a semidirect product of two abelian groups, $$H$$ is solvable.
• As a quotient of the solvable group $$H$$, $$G$$ is solvable. But a perfect solvable group must be trivial.

I don't expect to find a "conceptual" proof that $$c\in \langle a,b\rangle_G$$. But I recently learned a beautiful visual proof of this fact due to Josh Hinman (shared here with his permission).

The relations $$ab = b^2a$$, $$bc = c^2b$$, and $$ca = a^2c$$ can be represented by commutativity of the following diagrams.

Now we can paste together $$10$$ of these pentagonal diagrams to form a dodecahedron with two missing faces:

Tracing around the missing faces gives the relation $$c = b^{-1}ab^{-1}a^{-3}b$$, so $$c\in \langle a,b\rangle_G$$.